Question

Assuming no heat loss, what final temperature is expected after
50,0 g of water at 20.0°C is mixed with 30,0 g of water at 80.0°C?
(the specific heat of water is 4.18 J/Kg K)
04260
50.00
17.50​

Answer 1

Answer:

since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed.  Since the masses of the liquid at different temperatures are same, the answer is very easy and simple :  average of 20⁰C and 40⁰C.  that is:  30⁰C.

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final temperature of the mixture = 

     = [ m1 * T1 + m2 * T2 ] / (m1 + m2)

     = [ 50 gms * 20⁰ C + 50 gms * 40⁰C ]  / (50+50)

     = 3,000 / 100 = 30⁰C

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another way using specific heats :

   let the final temperature be = T ⁰C

   Amount of heat given out by the hot water = m * s * (40⁰C - T)

           = 50 gms * s* (40 -T)

   Amount of heat taken in by the cold water = m * s * (T - 20⁰C)

       = 50 gms * s * (T - 20 )

   As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,

           50 * s * (40 -T) = 50 gm * s  * (T-20) 

                 40 - T = T - 20

             2 T = 60    => T = 30⁰C